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There is a singularly perturbed problem:

\begin{align} &\epsilon u''(x) + u(x) - u^2(x)=0,\quad x<0<1,\\ & u(0)=1, \quad u(1)=0. \end{align}

where $\epsilon \ll 1$ is a tiny parameter. From the numerical result, it has a boundary layer at $x=1$ (The graph below is the numerical solution). I would like to find the asymptotic solution with the method of matched asymptotic expansions, but I have no idea how to deal with such a nonlinear problem.

Could anyone give me some hints? Of course, other methods or the exact solution are also available.

enter image description here

Here, I post the analytic result that I get until now (haven't finished yet). Consider the matched asymptotic expansion.

Step1: Outer Solution

Assume that the solution can be expanded in powers of $\epsilon$, namely, $$u(x,\epsilon)=u_0(x)+\epsilon u_1(x) + \mathcal{O}(\epsilon^2)$$ Substituting this into original problem and consider the $\mathcal{O}(1)$ equation, that is $\epsilon=0$, we get \begin{align} & u_0^2(x)-u_0(x)=0, \\ & u_0(0)=1. \end{align} We know that $u_0(x)=1$ is a or solution. From the graph above, it is reasonable.

Step2: Boundary Layer

Based on the numerical result, there is a boundary layer at $x=1$, so the stretching transformation we take is $$\bar x = \frac{x-1}{\delta}. $$

From the stretching transformation and the chain rule, we have that $$\frac{d}{dx}=\frac{1}{\delta}\frac{d}{d \bar x}.$$

Let the $U(\bar x)$ be the solution using the stretching transformation, then the original problem becomes \begin{align} &\frac{\epsilon}{\delta^2}U''(\bar x)+U(\bar x)-U^2(\bar x)=0, \\ &U(0)=0. \end{align} The boundary condition comes from the fact that $u(1)=U(\frac{1-1}{\delta})=U(0)=0.$ The domainant balances illustrates $\frac{\epsilon}{\delta^2}=1$, namely, $$\epsilon = \delta^2. $$ Thus, the transformed problem becomes \begin{align} &U''(\bar x)+U(\bar x)-U^2(\bar x)=0, \\ &U(0)=0. \end{align} After solving the equation above, we may get the solution that includes a parameter $c$. Then, matching the solution of boundary layer and outer solution may get the parameter $c$, and obtain the final asymptotic solution. But I have no idea how to solve the equation above.

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You get a center around $u=0$ and a saddle point around $u=1$. To get at least one-sided convergence the solution has to follow the level curve of the saddle point, $$ ϵu'^2+u^2-\frac23u^2=\frac13. $$ This gives $u'(1)=\pm\frac1{\sqrt{3ϵ}}$, giving an IVP for this curve. This solution will not exactly meet $u(0)=1$, only come very close to it, with an error of the form $e^{-c/\sqrtϵ}$ which is smaller than any power of $ϵ$ asymptotically for $ϵ\to0$.

This should however be good enough for a first approximation.

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  • $\begingroup$ Thanks for your answer. I drew the graph according to your answer, and the solution is $u(x)=sin(\frac{x-1}{\sqrt{3\epsilon}})$. It is indeed a good approximation. This problem, however, has several solutions. One of them is shown in the graph that I added above. So I would like to ask how can I get the asymptotic solution shown in the graph? sorry for the unclear question before. By the way, how can I get the level curve of the saddle point that you post? I study the numerical stuff and have little knowledge about the analytic one. $\endgroup$
    – ZR Tang
    Jun 15, 2021 at 12:38

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